Which of the following is/are correct?
A. \( \int e^x\left[\displaystyle\frac{1}{(x-2)}-\displaystyle\frac{1}{(x-2)^2}\right]dx=\displaystyle\frac{e^x}{(x-2)}+C \) , where C is an arbitrary constant.
B. \( \int e^x\left[\displaystyle\frac{1}{(x-2)^3}-\displaystyle\frac{3}{(x-2)^4}\right]dx=\displaystyle\frac{e^x}{(x-2)^2}+C \) , where C is an arbitrary constant.
C. \( \int e^x\left[\displaystyle\frac{x-4}{(x-2)^3}\right]dx=\displaystyle\frac{e^x}{(x-2)^3}+C \), where C is an arbitrary constant.
D. \( \int e^x\left[\displaystyle\frac{1}{x+2}-\displaystyle\frac{3}{(x+2)^2}\right]dx=\displaystyle\frac{e^x}{(x+2)^2}+C \) , where C is an arbitrary constant.
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → A only
The result \( \int e^x\big(f(x)+f'(x)\big)\,dx=e^x f(x)+C \) applies when the bracket is a function plus its derivative; the answer is then \( e^x \) times that function.
In A:
\( \displaystyle f(x)=\frac{1}{x-2},\quad f'(x)=-\frac{1}{(x-2)^2} \)
In B:
\( \displaystyle f(x)=\frac{1}{(x-2)^3},\quad f'(x)=-\frac{3}{(x-2)^4} \)
In C, writing x − 4 = (x − 2) − 2:
\( \displaystyle \frac{x-4}{(x-2)^3}=\frac{1}{(x-2)^2}-\frac{2}{(x-2)^3} \)
\( \displaystyle f(x)=\frac{1}{(x-2)^2},\quad f'(x)=-\frac{2}{(x-2)^3} \)
In D, differentiating the right side:
\( \displaystyle \frac{d}{dx}\left[\frac{e^x}{(x+2)^2}\right]=e^x\left[\frac{1}{(x+2)^2}-\frac{2}{(x+2)^3}\right] \)
So the integrals in A, B and C are \( \frac{e^x}{x-2}+C \), \( \frac{e^x}{(x-2)^3}+C \) and \( \frac{e^x}{(x-2)^2}+C \), and the derivative of \( \frac{e^x}{(x+2)^2} \) is \( e^x\left[\frac{1}{(x+2)^2}-\frac{2}{(x+2)^3}\right] \).
Checking the statements
(A) \( \int e^x\left[\frac{1}{(x-2)}-\frac{1}{(x-2)^2}\right]dx=\frac{e^x}{(x-2)}+C \) — Correct. The bracket is \( \frac{1}{x-2} \) plus its derivative.
(B) \( \int e^x\left[\frac{1}{(x-2)^3}-\frac{3}{(x-2)^4}\right]dx=\frac{e^x}{(x-2)^2}+C \) — Incorrect. The integral is \( \frac{e^x}{(x-2)^3}+C \).
(C) \( \int e^x\left[\frac{x-4}{(x-2)^3}\right]dx=\frac{e^x}{(x-2)^3}+C \) — Incorrect. The integral is \( \frac{e^x}{(x-2)^2}+C \).
(D) \( \int e^x\left[\frac{1}{x+2}-\frac{3}{(x+2)^2}\right]dx=\frac{e^x}{(x+2)^2}+C \) — Incorrect. The derivative of the right side is \( e^x\left[\frac{1}{(x+2)^2}-\frac{2}{(x+2)^3}\right] \), not the given integrand.